How do you solve Minimum Sum of Values by Dividing Array on LeetCode?

Minimum Sum of Values by Dividing Array (LeetCode #3117) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Understanding bitwise operations is crucial for solving this problem.

What is the brute force solution for Minimum Sum of Values by Dividing Array?

The brute force approach for Minimum Sum of Values by Dividing Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible way to partition the array into contiguous subarrays and calculating their AND values. This method guarantees that we will find a solution if one exists, but it is inefficient due to the large number of possible partitions. The optimal Optimal Solution approach improves this to O(n * m).

What are the interview tips for Minimum Sum of Values by Dividing Array?

Always clarify the problem requirements before diving into coding. Think about edge cases and how they might affect your solution. Discuss your thought process with the interviewer to demonstrate your understanding.

What are common mistakes when solving Minimum Sum of Values by Dividing Array?

Not considering all possible partitions, leading to incorrect results. Failing to handle edge cases where partitions cannot be formed.

#3117

Minimum Sum of Values by Dividing Array

Hard

ArrayBinary SearchDynamic ProgrammingBit ManipulationSegment TreeQueueDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n * m)

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Intuition

Time O(n * m)Space O(n * m)

The optimal solution uses dynamic programming to efficiently calculate the minimum sum by storing results of subproblems. This avoids redundant calculations and allows us to build up the solution incrementally.

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Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i][j] represents the minimum sum of the last elements of the first j subarrays formed from the first i elements of nums.
2Step 2: Iterate through nums and for each element, check possible subarrays that can end at that element and match the corresponding andValue.
3Step 3: Update the DP array based on the valid partitions found, ensuring to keep track of the minimum sum.

solution.py20 lines

1# Full working Python code
2import sys
3
4
5def min_sum_partition(nums, andValues):
6    n = len(nums)
7    m = len(andValues)
8    dp = [[sys.maxsize] * (m + 1) for _ in range(n + 1)]
9    dp[0][0] = 0
10    
11    for j in range(1, m + 1):
12        for i in range(1, n + 1):
13            and_value = nums[i - 1]
14            for z in range(i, 0, -1):
15                and_value &= nums[z - 1]
16                if and_value == andValues[j - 1]:
17                    dp[i][j] = min(dp[i][j], dp[z - 1][j - 1] + nums[i - 1])
18                if and_value == 0:
19                    break
20    return dp[n][m] if dp[n][m] != sys.maxsize else -1

ℹ

Complexity note: The time complexity is O(n * m) because we iterate through each element of nums and for each element, we check possible subarrays for each value in andValues.

1Understanding bitwise operations is crucial for solving this problem.
2Dynamic programming can significantly reduce the complexity of problems involving partitions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.